Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)
\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)
\(x\times\left(8-25\right)=2\times29-33\)
\(x\times-17=25\)
\(x=-\dfrac{25}{17}\)
2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)
\(15\div\left(x+2\right)=\left(27+3\right)\div1\)
\(15\div\left(x+2\right)=30\div1\)
\(15\div\left(x+2\right)=30\)
\(x+2=\dfrac{1}{2}\)
\(x=-\dfrac{3}{2}\)
3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)
\(20\div\left(x+1\right)=\left(25+1\right)\div13\)
\(20\div\left(x+1\right)=26\div13\)
\(20\div\left(x+1\right)=2\)
\(x+1=20\div2\)
\(x+1=10\)
\(x=9\)
4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)
\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)
\(320\div\left(x-1\right)=100\div4+15\)
\(320\div\left(x-1\right)=25+15\)
\(320\div\left(x-1\right)=40\)
\(x-1=8\)
\(x=9\)
5) \(240\div\left(x-5\right)=2^2\times5^2-20\)
\(240\div\left(x-5\right)=4\times25-20\)
\(240\div\left(x-5\right)=100-20\)
\(240\div\left(x-5\right)=80\)
\(x-5=30\)
\(x=35\)
6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)
\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)
\(70\div\left(x-3\right)=80\div4-10\)
\(70\div\left(x-3\right)=20-10\)
\(70\div\left(x-3\right)=10\)
\(x-3=7\)
\(x=10\)
a, \(x\) + 99: 3 = 55
\(x\) + 33 = 55
\(x\) = 55 - 33
\(x\) = 22
b, (\(x\) - 25):15 = 20
\(x\) - 25 = 20 x 15
\(x\) - 25 = 300
\(x\) = 300 + 25
\(x\) = 325
c, (3\(x\) - 15).7 = 42
3\(x\) - 15 = 42:7
3\(x\) - 15 = 6
3\(x\) = 6 + 15
3\(x\) = 21
\(x\) = 21: 3
\(x\) = 7
\(1,\frac{2}{3}+\frac{4}{9}+\frac{1}{5}+\frac{2}{15}+\frac{3}{2}-\frac{17}{18}\)
\(< =>\frac{4}{9}+\frac{3}{2}+\left(\frac{2}{3}+\frac{1}{5}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{8}{18}+\frac{27}{18}+\left(\frac{10}{15}+\frac{3}{15}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{35}{18}+1-\frac{17}{18}\)
\(< =>\frac{53}{18}-\frac{17}{18}\)
\(< =>2\)
\(2,\frac{13}{28}\cdot\frac{5}{12}-\frac{5}{28}\cdot\frac{1}{12}\)
\(< =>\left(\frac{13}{28}-\frac{5}{28}\right)\cdot\left(\frac{5}{12}-\frac{1}{12}\right)\)
\(< =>\frac{2}{7}\cdot\frac{1}{3}\)
\(< =>\frac{2}{21}\)
\(3,\frac{19}{4}\cdot\frac{15}{23}-\frac{15}{4}\cdot\frac{7}{23}+\frac{15}{4}\cdot\frac{11}{23}\)
\(< =>\frac{285}{92}-\frac{105}{92}+\frac{165}{92}\)
\(< =>\frac{15}{4}\)
a) \(-28-7|-3x+15|=-70\)
\(\Rightarrow7|-3x+15|=42\)
\(\Rightarrow|-3x+15|=6\)
\(\Rightarrow|3\left(5-x\right)|=6\)
\(\Rightarrow|3|.|5-x|=6\)
\(\Rightarrow3|5-x|=6\)
\(\Rightarrow|5-x|=2\)
\(\Rightarrow\orbr{\begin{cases}5-x=2\\5-x=-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}\)
Vậy \(x\in\left\{3;7\right\}\)
b) \(|18-2|-x+5||=12\)
\(\Rightarrow\orbr{\begin{cases}18-2|-x+5|=12\\18-2|-x+5|=-12\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2|5-x|=6\\2|5-x|=30\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}|5-x|=3\left(1\right)\\|5-x|=15\left(2\right)\end{cases}}\)
Từ \(\left(1\right):\Rightarrow\orbr{\begin{cases}5-x=3\\5-x=-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=8\end{cases}}\)
Từ \(\left(2\right):\Rightarrow\orbr{\begin{cases}5-x=15\\5-x=-15\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-10\\x=20\end{cases}}\)
Vậy \(x\in\left\{2;8;-10;20\right\}\)
c) \(12-2\left(-x+3\right)^2=-38\)
\(\Rightarrow2\left(3-x\right)^2=50\)
\(\Rightarrow\left(3-x\right)^2=100\)
\(\Rightarrow\orbr{\begin{cases}3-x=10\\3-x=-10\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-7\\x=13\end{cases}}\)
Vậy \(x\in\left\{-7;13\right\}\)
d) \(-20+3\left(2x+1\right)^3=-101\)
\(\Rightarrow3\left(2x+1\right)^3=-81\)
\(\Rightarrow\left(2x+1\right)^3=-27\)
\(\Rightarrow2x+1=-3\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
Trả lời:
a, -28 - 7| -3x + 15 | = -70
=> 7| -3x + 15 | = 42
=> | -3x + 15 | = 6
=> -3x + 15 = 6 hoặc -3x + 15 = -6
=> -3x = -9 -3x = -21
=> x = 3 x = 7
Vậy x = 3; x = 7
b, | 18 - 2 | -x + 5 || = 12
=> 18 - 2| -x + 5 | = 12 hoặc 18 - 2| -x + 5 | = -12
=> 2 | -x + 5 | = 6 hoặc 2 | -x + 5 | = 30
=> | -x + 5 | = 3 hoặc | -x + 5 | = 15
=> -x + 5 = 3 hoặc -x + 5 = -3 hoặc -x + 5 = 15 hoặc -x + 5 = -15
=> x = 2 x = 8 x = -10 x = 20
Vậy x \(\in\){ 2; 8; -10; 20 }
c, 12 - 2.( -x + 3 )2 = -38
=> 2.( -x + 3 )2 = 50
=> ( -x + 3 )2 = 25
=> -x + 3 = 5 hoặc -x + 3 = -5
=> x = -2 x = 8
Vậy x = -2; x = 8
d, -20 + 3.( 2x + 1 )3 = -101
=> 3.( 2x + 1)3 = -81
=> ( 2x + 1 )3 = -27
=> 2x + 1 = -3
=> 2x = -4
=> x = -2
Vậy x = -2
=> x = 1
a) 4 . x + 15 = - 5
⇒ 4 . x = - 5 – 15
⇒ 4 . x = - 20
⇒ x = - 20 : 4
⇒ x = - 5.
Vậy x = - 5.
b) (- 270) : x – 20 = 70.
⇒ (- 270) : x = 70 + 20
⇒ (- 270) : x = 90
⇒ x = (- 270) : 90
⇒ x = - 3
Vậy x = - 3
\(a.13.\left(78+22\right)=13.100=1300\)
\(b.25.70+28.25+25.2=25.\left(70+28+2\right)=25.100=2500\)
\(c.\dfrac{15.3.14.2.19}{19.14.15}=6\)
\(d.\dfrac{132.70.60}{12.35.20}=\dfrac{11.12.35.2.20.3}{12.35.20}=66\)
\(\frac{15}{4}+\frac{15}{28}+\frac{15}{70}+...+\frac{15}{x.\left(x+3\right)}=\frac{99}{20}\)
=> \(5.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{99}{20}\)
=> \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{99}{20}:5\)
=> \(1-\frac{1}{x+3}=\frac{99}{20}.\frac{1}{5}=\frac{99}{100}\)
=> \(\frac{1}{x+3}=1-\frac{99}{100}\)
=> \(\frac{1}{x+3}=\frac{1}{100}\)
=> x + 3 = 100
=> x = 100 - 3
=> x = 97